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Ch 3 Stoichiometry Atomic Masses Atomic masses are assigned relative to 12C which has exactly 12 amu Mass spectrophotometry is used to accurately determine the relative mass of atoms (see p. 80 for description) Average Atomic Mass Elements occur in nature as a mixture of isotopes. Average Atomic Mass – weighted average of the masses of the naturally occurring isotopes of an element Atomic Mass Example In nature carbon is composed of 98.89% 12C atoms and 1.11% 13C atoms. 12C has a mass of 12 amu and 13C has a mass of 13.0034 amu. What is the average atomic mass of carbon? Mass C = (0.9889)(12 amu) + (0.0111)*(13.0034 amu) Mass C = 12.01 amu The Mole Mole – number of carbon atoms in exactly 12 g of pure 12C 1 mole = 6.022 x 1023 units of a substance Avogadro’s Number = 6.022 x 1023 1 mole = atomic mass of an atom in grams Molar Mass Molar Mass – mass in grams of 1 mole of the compound EXAMPLE: Calculate the molar mass of H2O. H – 1.008 g/mol O – 16.00 g/mol Molar mass of H2O = (2(1.008) + 16.00) g/mol Molar mass of H2O = 18.02 g/mol Percent Composition Mass Percent = (mass of 1 mole of element in compound/ mass of 1 mole of compound) * 100% Mass percent of each component of a compound should add up to 100% Percent Composition Example What is the mass percent of carbon in glucose, C6H12O6. Mass Percent = (mass of 1 mole of element in compound)/ (mass of 1 mole of compound) * 100% Molar mass C = 12.01 g/mol Molar mass C6H12O6 = (6(12.01) + 12(1.008) + 6(16.00)) g/mol Molar mass C6H12O6 = 180.16 g/mol Mass percent C = ((6*12.01 g/mol)/(180.16 g/mol))*100% Mass percent C = 40.00 % Determining the Formula of a Compound Empirical Formula – a formula with the lowest whole-number ratio of elements in a compound Molecular Formula = (empirical formula)n where n is an integer Determining the Formula of a Compound Example A white powder is found to contain 43.64% phosphorus and 56.36% oxygen by mass. The compound has a molar mass of 283.88 g/mol. What are the compound’s empirical and molecular formulas? In a 100.00 g sample: 43.64 g P (1 mol P/30.97 g P) = 1.409 mol P 56.36 g O (1 mol O/16.00 g O) = 3.523 mol O Divide both by smaller value to get 1.409/1.409 = 1 P and 3.523/1.409 = 2.5 O Multiply both by 2 to get whole numbers to get an empirical formula of 2 P and 5 O or P2O5 with a molar mass of 141.94 g/mol Since molar mass of compound was 283.88 g/mol, which is twice as much as 141.94, the molecular formula is (P2O5)2 or P4O10 Chemical Equations Chemical equation – an expression representing a chemical reaction; the formulas of the reactants (on the left) are connected by an arrow with the products (on the right); states of matter are also given Example: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) Reactants Yield Products Balancing Chemical Equations Chemical equations must adhere to the Law of Conservation of Mass Whole number coefficients are added in front of chemical formulas to balance atoms on each side Chemical formulas are never changed! Balancing Chemical Equations Example Balance the following chemical equation: C2H5OH(l) + O2(g) → CO2(g) + H2O(g) Reactants – 2 C, 6 H, 3 O Products – 1 C, 2 H, 3 O Add coefficients: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g) Reactants – 2 C, 6 H, 7 O Products – 2C, 6 H, 7 O Mass is conserved and the equation is balanced. Stoichiometric Calculations Stoichiometry – portion of chemistry dealing with numerical relationships in chemical reactions; the calculation of quantities of a substances involved in chemical equations A means to determine quantities of reactants or products based on the amount of other species present in a reaction Mole Ratio Mole ratio – factor used in stoichiometry to convert moles of one substance to moles of another substance using the coefficients from the balanced chemical equation Steps Used in Stoichiometry Step 1: balance the equation for the reaction Step 2: convert known mass of reactants or products into moles of that substance Step 3: use the balanced chemical equation to set up the appropriate mole ratios Step 4: use the appropriate mole ratio to calculate the number of moles of the desired reactant or product Step 5: convert moles to proper units required by problem Stoichiometry Example 5.00 grams of solid magnesium reacts with hydrochloric acid to produce hydrogen gas and magnesium chloride. How many grams of hydrochloric acid must be present to ensure the complete reaction of magnesium in this reaction? Solution Write balanced chemical equation: Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq) Convert known value to moles: 5.00 g Mg (1 mol Mg/24.31 g) = 0.206 mol Mg Set-up appropriate mole-ratio and find desired moles: 0.206 mol Mg (2 mol HCl/1 mol Mg) = 0.412 mol HCl Covert to proper units required by problem: 0.412 mol HCl (36.46 g HCl/1mol HCl) = 15.0 g HCl Limiting Reactant Limiting reactant – reactant that is consumed first in a chemical reaction The amount of product formed is limited when one reactant is completely consumed before the others It is essential to determine which reactant is the limiting reactant when using stoichiometry to determine amount of products formed in a reaction How is a Limiting Reactant Calculated? Step 1: write and balance the equation for the reaction Step 2: determine moles of reactants from the known masses Step 3: use mole ratio to determine which reactant is limiting Limiting reactant is then used to calculate moles of product formed. Limiting Reactant Example 10.00 g of O2 reacts with 20.00 g of Cu metal to form copper (II) oxide. What is the limiting reactant? Write balanced chemical equation: O2(g) + 2Cu(s) → 2CuO (s) Determine moles of reactant: 10.00 g O2 (1 mol O2/32.00 g O2) = 0.3125 mol O2 20.00 g Cu (1 mol Cu/63.55 g Cu) = 0.3147 mol Cu Use mole ratio to determine which reactant is limiting: 0.3125 mol O2(2 mol Cu/1 mol O2) = 0.625 mol Cu 0.625 mol Cu>0.3147 mol Cu and Cu is limiting